Q) $x^{2}+x+1$ is a factor of $(x+1)^{n} - x^{n} - 1$, whenever
(A) $n$ is odd
(B) $n$ is odd and a multiple of $3$
(C) $n$ is an even multiple of $3$
(D) $n$ is odd and not a multiple of $3$
My Attempt :- I don't know the formal approach to prove it. I have taken some small values of $n$ and I have eliminated options (A),(B) and (C) as :-
Here, I can rewrite $(x+1)^{n} - x^{n} - 1$ as :-
$[\binom{n}{0}x^{n} + \binom{n}{1}x^{n-1} + \binom{n}{2}x^{n-2} +......+\binom{n}{r}x^{n-r}+.....+\binom{n}{n-1}x^{1} + \binom{n}{n}x^{0}] – x^{n} - 1$
$\binom{n}{1}x^{n-1} + \binom{n}{2}x^{n-2} +......+\binom{n}{r}x^{n-r}+.....+\binom{n}{n-1}x^{1}$
Now, if $n=3$ then it becomes $3x^{2} + 3x$. Since , $x^{2}+x+1$ does not divide it. So, $x^{2}+x+1$ is not a factor of $(x+1)^{n} - x^{n} - 1$ when $n=3$ , So, I have eliminated options (A) and (B)
Now, when $n=6$ then it becomes $\binom{5}{1}x^{5} + \binom{6}{2}x^{4}+\binom{6}{3}x^{3}+ \binom{6}{4}x^{2}+\binom{6}{5}x^{1}$ $=6x^{3}(1+x+x^{2}) + 6x(1+x+x^{2}) + 9x^{4}+8x^{3}+9x^{2}$.So, $x^{2}+x+1$ is a not a factor of $(x+1)^{n} - x^{n} - 1$ when $n=6$. So, I have eliminated option $(C)$.
So, I am getting answer as $(D)$ but I don't know how to prove it formally for all $n$=odd which should also not be multiple of $3$. Please help.